Q91 / 160 · WBJEE 2011 · Mathematics

Let \(\mathrm{a}, \mathrm{b}, \mathrm{c}\) be three real numbers such that \(\mathrm{a}+2 \mathrm{~b}+4 \mathrm{c}=0\). Then the equation \(\mathrm{ax}^2+\mathrm{bx}+\mathrm{c}=0\)

  • A
    has both the roots complex
  • B
    hat its roots lying within \(-1 < x < 0\)
  • Correct answer
    C
    has one of roots equal to \(\frac{1}{2}\)
  • D
    has its roots lying within \(2 < x < 6\)

Solution

Hints: \(\frac{1}{4} \mathrm{a}+\frac{1}{2} \mathrm{~b}+\mathrm{c}=0\) \(\begin{aligned} & \left(\frac{1}{2}\right)^2 a+\left(\frac{1}{2}\right) b+c=0 \\ & \therefore x=\frac{1}{2} \end{aligned}\)

Hence, the correct answer is option: 'C'.

Relevant Questions